TRANSACTIONS
OF THE
AMERICANMATHEMATICALSOCIETY
Volume 192,1974
FIELDS OF FRACTIONSFOR GROUP ALGEBRASOF FREE GROUPS
BY
JACQUESLEWIN(l)
ABSTRACT. Let KF be the group algebra over the commutative field K of the free
group F. It is proved that the field generated by KF in any Mal'cev-Neumann embedding
for KF is the universal field of fractions U(KF) of KF. Some consequences are noted. An
example is constructed of an embedding KF C D into a field D with D * U(KF). It is
also proved that the generalizedfree product of two free groups can be embedded in a field.
I. Introduction. P. M. Cohn has recently shown [3, Chapter 7] that if R is a
semifir then there is an embedding of R in a (not necessarily commutative) field
U(R) which is universal in the sense that if R C D is another embedding of R in
a field then there is a specialization of U(R) onto D which extends the identity
map of R. In particular, free associative algebras and free group algebras have
universal fields of fractions.
Let now F be a free group and K a commutative field. I. Hughes [5] singles out
a class of "free" embeddings (see definition below, §11) of KF into fields and
shows that any two free embeddings which are both generated (qua fields) by KF
are ^TF-isomorphicThis makes it plausible that U(KF) is a free embedding and
we show that this is indeed the case. Oddly enough this is not proved by verifying
directly the freedom property of U(KF), but by first proving a subgroup
theorem: If G is a subgroup of F, then KG generates, in U(KF), its universal field
U(KG).
The significance of our theorem is that it shows that U(KF) is in fact the field
generated by KF in any Mal'cev-Neumann embedding of KF [10]. If R is a free
tf-algebra on the generators of F, then it is easily seen that U(R) = U(KF). Thus
we have both U(R) and U(KF) represented in power series over F. This has
several interesting consequences: U(F) can be ordered; the center of U(F) is K(if
F is not commutative); there is a homomorphism of the multiplicative group
U(F)* onto the free group F. (F is actually a retract of U(F)*.)
Going back to groups, we show that any generalized free product G of two free
groups can be embedded in a field. However, using an example of M. Dunwoody
[4], we show that there need not exist a fully inverting (definition below)
embedding for G, even if the amalgamated subgroup is cyclic.
Hughes [5] asks whether there exists a nonfree embedding of the free group
algebra KF. We close by exhibiting such an embedding.
Receivedby the editors December 11, 1972.
AMS(MOS)subjectclassifications
(1970).Primary16A26,20E05,16A06.
Key words and phrases, free algebras, free group algebras, universal fields of fractions,
generalized free products of free groups.
(') This workwas supportedby NSF grant GP 33050.
CopyrightO 1974,AmericanMathematicalSociety
339
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
340
JACQUESLEWIN
II. U(KF) is Hughes-free. If F is a semifir, we denote by U(R) the universal
field of fractions of R. R Q U(R) is fully inverting: every full F-matrix inverts
over U(R). Every element m, of U(R) is rational over R, i.e. «, is the first
component of a solution u of a matrix equation Au + a = 0 where A is a full
nXn F-matrix and a G F". (Recall that A is full if A is not a product of two
matrices of smaller size.) If S is a subring of F, the inclusion S -> Ris honest if
every full 5-matrix is still full as an F-matrix.
If h: F -» D is a homomorphism into a field D, then there is universal
specialization/?: U(R) -* D which extends h. The domain ofp consists of the set
of entries of inverses of those F-matrices whose image is invertible over D.
Details and proofs may be found in Cohn [3, Chapter 7].
In particular, if F is a free group and K a commutative field, then KF is a
semifir so the above results apply. We write U(F) for U(KF).
If H is a sugroup of Fand FF is embedded in a field D, we denote by DivD(H)
the smallest subfield of D which contains H and K. Note that DivD(H) is rational
over H.
Our aim is to show that for any subgroup G of a free group F, Di\^(G)
= U(G).
The universal specialization from U(G) into U(F) will be a monomorphism
U(G) -* U(F) provided the inclusion KF -* KG is honest. It is this that we shall
prove. We first deal with a special case.
Lemma 1. Let F be a free group and G a normal subgroup of finite index in F.
Then the inclusionKG -* KF is honest.
Proof. Let sx = 1, i2, ..., s„ be a set of coset representatives for G in F. Then
KF = ®,"=i(FG)i,. Right multiplication by an element of KF is a left KF, and
hence a left KG, module homomorphism. Thus we have a faithful representation
<p:KF -» (KG)n, the n X n matrices over KG.
If v E KG, then s,v = SjVsf1• s,, = v" • s¡. Since G is normal in F, v'1 E KG.
Thus wpis the diagonal matrix
vs'
(O
°1
v<p=
0
v1']
In the obvious way, we extend <pto matrices over KF. Let now M be a FG-matrix
which is full over KG. Since conjugation by s¡ is an automorphism of KG, Ms' is
also full over KG for i = 1, ..., n. KG is again a free group algebra, so KG is a
fir. It follows[3,Theorem 6.4, p. 282]that the diagonal sum N = M + M'1 + • • •
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
FIELDS OF FRACTIONSFOR GROUP ALGEBRAS
341
+ Ms" is full. But, by (1), Mtp is similar, via a permutation matrix, to N. So M, as
a Â'F-matrix, maps under tp to a full matrix. So M is full over KF and the lemma
is proved.
The next step is to drop the normality assumption on G. We first need an easy
lemma.
If 5 is a subset of a AT-algebra,denote by AT<5>the subalgebra it generates.
Lemma 2. Let G be a normal subgroup offinite index in the group F, and suppose
KF is embeddedin afield D. Then Divß(F) has finite dimensionas a left DivD(G)
vector space. Further Divß(F) = K(X>i\D(G),F}.
Proof. Let i, = 1, s2,..., sn be a set of coset representatives for G in F, and
consider A = 0f_, DivD(G)s¡. Now s¡ induces, by conjugation, an automorphism
of DivD(F) which leaves G, and hence DivDG,invariant. Thus for d E Divfl(G),
d'1 E Drv¿(G). Thus since s¡d = d''s¡ and s,Sj = gySk for some k and some
gy E G, A is a ^-algebra. Since A has finite left Divfl(G) dimension, and A is an
integral domain, A is a field. Since A contains F, A = Div^F). Clearly A is
generated by DivD(G)and F. O
Lemma 3. Let F be a free group, and L a subgroup of finite index in F. Then
DivuiF)(L) = U(L).
Proof. Let G be the intersection of the conjugates of L. Then G is still of finite
index in Fand is of course normal. Then, by Lemma 1, Div^^^G) = U(G) and
Div^iG)
= U(G).
Let p be the universal specialization p: U(L) -» Div^^L). Since a full KGmatrix inverts over U(G) it inverts over both U(L) and DiVy^L). So the entries
of inverses of full ^TG-matricesare in the domain £ of p. So U(G ) Q £. Since p
is an ¿-specialization, also L Q L So, by Lemma 2, U(L) = K(U(G),L) C £.
Since p is onto Di\v^ L,p is an ¿-isomorphism Div^L ^ U(L).
The next two lemmas allow us to go from subgroups of finite index to arbitrary
finitely generated subgroups.
Lemma 4. Suppose the free group F is the free product Hx * H2 of two subgroups.
ThenDivu(F)(Hx)= U(HX).
Proof (P. M. Cohn). Embed KHX*KKH2in U(HX)*KU(H2).If M is a KHXmatrix which is full over KHX,then M inverts over U(HX) and hence over
U(HX)* U(H2).So M is full over KHX* KH2 = KF. So M inverts over U(F) and
hence inverts over Div^ (//,). D
Lemma 5. Let H be a finitely generated subgroup of the free group F. Then H is
a free factor in a subgroup L of finite index in F. D
A proof may be found in [6].
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
342
JACQUESLEWIN
Theorem 1. If H is a subgroup of the free group F, then Dwu(F)(H) = U(H).
Proof. If H is finitely generated, this is the contents of Lemmas 3, 4 and 5. Let
now H be an arbitrary subgroup of F, and let M be a F//-matrix which is full
over KH. Then there is a finitely generated subgroup //' of H such that M is a
matrix over KH'. M is still full over KH'. Thus, by the finitely generated case, M
inverts over U(F), and hence over Dity^//).
□
Recall that a group G is indexed at (H, t) if H is normal in G and G/H is the
infinite cyclic group generated by the coset tH. Let F be a free group and
KF C D an embedding of KF in a field. I. Hughes [5] makes the following
definition: the embedding KF C D is free if for any finitely generated subgroup
G of F, and indexing (H,t) of G, then the powers of t are left DivD(KH)independent. He then shows
Hughes' theorem. If KF C Dx,KF C D2 are twofree embeddingsof KF then
there is a KF-isomorphismDivDi(KF) « Divß2(KF). D
Proposition 6. KF C U(F) is a Hughes-freeembedding.
Proof. Let G be a finitely generated subgroup of F, (H, t) an indexing of G.
Since H is normal in G, conjugation by t induces an automorphism t of
U(H) ç U(F). Form the skew Laurent polynomial ring F = U(H)[z,z~x] with
the commutation rule dz = z(dr). Then F is an Ore domain with quotient field
D. Since P is a principal ideal domain, the embedding P C D is fully inverting
so that D = U(P).
Now we have a homomorphism tp: P -+ K(U(H), t,t~x > which maps z to r and
hence, composing with the inclusion K(JJ(H), t, t~x) -* U(G), a homomorphism
h: P -> U(G).
U(G) %----?.i/(P)
ui
K<U(H),t, rl)
^SvȒL
«-—
Ul
U(H) <-!-»
ul
p
Ul
i/(//)
A then extends to a specialization p: U(P) -* U(G). But, identifying KG
= K(H,t,rxy with K(H,z,z~1}, p is a FG-specialization. Since i/(G) is fully
inverting for FG, so is i/(F) (Cohn [3, Theorem 2.3, p. 257]).Hence U(G)and
U(P) are KG isomorphic. Clearly, a FG-isomorphism is the identity on H, and
hence maps U(H) onto itself. Since {z1}is left U(H) independent, {/'} is left U(H)
independent and the proposition is proved.
III. A series representation and applications. We first note that if F = K(X) is
the free F-algebra on a set X and F is the free group on X, then U(R) = U(F);
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
FIELDS OF FRACTIONSFOR GROUP ALGEBRAS
343
since x E X has an inverse in U(R), there is a homomorphism KF'-* U(R)
which is the identity on R. This homomorphism extends to a specialization
p: U(F) -» U(R). So every matrix over R which inverts over U(R) inverts over
U(F), i.e. every full P-matrix inverts over U(F). Since U(F) is generated by R, p
is an isomorphism.
Recall the Mal'cev-Neumann method for embedding a free group algebra KF
in a field [10]: order F and let D be the set of formal series over F, with
coefficients in K, whose support is well ordered. If 0 # p E D, then p can be
written uniquely as p — kf(I + p'), k E K, f E F, p' = 0 or p1 with positive
support. Then p~x = (1 —p' + p'2 — • • • )g~xk~x. Thus if H is a subgroup of F,
Div^//) consists of power series whose support is in H. This makes it clear that
D is Hughes-free. Thus, applying Hughes' theorem and the proposition, we
obtain
Theorem 2. Let F be a free group on the set X, R = K(xy the free K-algebra on
A,-D any Mal'cev-Neumannembeddingof F. Then U(R) = U(F) s± Divfl(F). D
Theorem 3. If R is a free algebra over the ordered (commutative)field K, then
U(R) can be ordered.
Proof. We need only note that a Mal'cev-Neumann field can be ordered if K
can be. D
Theorem 4 (cf. Klein [8]). If Risa noncommutativefree K-algebra, then the center
of U(R)is K.
Proof. Let R be freely generated by A, | A| > 1, and let F be the free group on
A. We may consider U(R) as embedded in a Mal'cev-Neumann field for F. Let
z = 2/ef kff be in the center of U(R), and say /, is the least element in the
support of z. Then, for x E X, xfx and/,x are the least elements in the supports
of xz and zx. So/, is in the center of F, i.e./, = 1. So kx —z is again in the center
of U(R). But then kx —z = 0 or its support consists of positive elements. This
last leads to a contradiction and hence z E K.f2
Theorem 5. Let R be a free K-algebra, F the corresponding free group, and let
U(R)* be the multiplicative group of nonzero elements of U(R). Then the free group
F is a retract of U(R)*.
Proof. We regard U(R) as embedded in a Mal'cev-Neumann field for F. Let N
be the subset of U(R)* of elements k + P, where 0 ¥=k E K, and P = 0 or P
has positive support. An easy calculation shows that N is normal in U(R)*. If
gx,g2 are different elements of F, then gxg2x # 1. So gxg2x E N and thus
g, # g2modW. Also, if ß E U(R)*, Q can be written uniquely as Q =
gk(l + Q') with¿(1 + Q') E N, g E F. Then ß = g mod N and ß -» g is the
required retraction of U(R)* onto F. D
Corollary. Let U(R)*i,be the commutator factor group of U(R)*. The projection
U(R)* -* U(R)*i, is injective on the generators of R. D
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
344
JACQUESLEWIN
(This provides a partial answer to problem 10 on p. 286 of [3].)
IV. Generalized free products of free groups. Recall from [3] the following
fundamental property of free products of rings over a (skew) field D. Let F,, F2
be D-rings and let {1}U S¡ be a left Z)-basis for F,. Then the monomials on
Sx U S2, no two successive letters of which are in the same factor, form, together
with 1, a left Z)-basisfor the free product F, *DR2.
Theorem 6 (cf. [1, Corollary 3.1], [7, Theorem 9], [9]). Let Fx,F2 be twofree
groups with a common subgroup H and let G be the generalized free product Fx*HF2.
Then the group algebra KG can be embedded in afield.
Proof. If {1} U S¡ is a left transversal for Hin F¡ then it is clear by looking in
Mal'cev-Neumann fields that {1} U S¡ is a left Dity^/O-independent
set.
Further there are KH isomorphisms Divv^(H) ^ Divy^H) ^ {/(//). These
observations and the remark preceding the theorem show that the free product
R = U(FX)*u(h) V{F2)makes sense and embeds KG. But F is a free ideal ring [2]
and so has a universal field of fractions /7(F). Thus KG Q U(R). D
Unfortunately, U(R) need not be fully inverting for KG. Indeed KG need not
have a fully inverting embedding. For Dunwoody [4] has shown that if
G = (a,b; a2 = ¿>3>,then KG has a nonfree finitely generated projective module
P. Such a ring has a full matrix which is not invertible in any field which embeds
it: let M be a free module of least rank such that M = P, 8 P. The projection
M -* Pis given by an idempotent matrix u which is not the identity. Thus /i does
not become invertible in any overfield. However, u is full. For otherwise P Q N,
a submodule of M with fewer generators. Since P is a direct summand of M, it
is a direct summand of N, and hence needs fewer generators than M, contradicting the minimality of the rank of M.
V. An example. We now construct a nonfree embedding of a free group algebra
in a field.
Let F be the free group Fx * F2 where Fxis free on z and F2 is free on x and w.
We embed kF¡ in F, = U(Ft). In F, we choose a F-basis {1}U 5, such that
{(1 +z)_1,z'';/' = ±1,±2,...} C S, and in F2 we choose a F-basis {1}U S2
with F2\{1) C S2.Let b = (1 + z)~x.In F = F, *KR2let T be the set of elements
r = f\bf2b • • -fnbfn-i where / G F is a reduced word which neither begins or
ends with z±x for /' = 2,..., n — 1,/, = 1 or/, does not end in z±l,f2 = 1 or/2
does not begin with z±x.We extend the length function / of F to a length function
on T = T U F\{1} by declaring1(f)= n + 2,"-! Kf¡)-It is clear that a set of
elements of T that are spelled differently is left F-independent. We embed F in
its universal field U(R). In U(R) we may choose a basis {1} U S with T C S. Let
u = bw(\ + x), and let Q = Divu(R)(k[u]).We claim that in U(R) the set Fis left
(¿-independent.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
FIELDS OF FRACTIONSFOR GROUP ALGEBRAS
345
We first note that ß is the field of right quotients of k[u] so that a set is left Qindependent if and only if it is &[u]-independent. So we need only show that F is
left £[t/]-independent. We note next that bw and bwx freely generate a free
subalgebra of U(R). Let then fa be distinct elements of F and suppose that there
are polynomials pa(u), not all zero with 2 Pa(u)fa = 0. Choose a such that pa(u)
has maximal u degree, say n, and fa has maximal length among the fß for which
Pß(ü) has degree n. Two cases arise.
1. fa does not begin with x_1. Then pa(u)f„ gives rise to a term t
= (bwx)"~xbwxfa.This term has length 4m + l(fa) and has degree n on b. Also it
is clear that this length is maximal among the monomials in the expansion of
2 Pa(u)fa which have degree n on b. So since the sum vanishes, for some
ß ¥= a, Pß(ü) has degree n and the term t also appears in the expansion of u"fß.
Since fa had maximal length, this forces fa = fß, a contradiction.
2. We may then assume thatjf starts with x_1. Nowpa(«)/, gives rise to a term
(bwx)"~xbwfaof length An + l(fa) — 1, and this is the only term of this length in
the expansion of pa(u)fa (since all other terms in the expansion of pa(u)fa either
end with x or are too short). It is again easy to see that this implies that fß=fa
for some a ¥= ß, and this contradiction proves the claim.
We now provide ourselves with another copy U(R)' of U(R) and consider the
free product V = U(R) * U(R)' amalgamating ß with ß'. Then the group words
on the letters z, x, w, z', x', w' are left ß-independent. and hence ^-independent.
Thus the group G generated by these letters is free on them and the AT-algebra
generated by G in U(V) is the group algebra KG. Now V is still a free ideal ring
[2] and hence we may embed V in U(V). Clearly G generates U(V) qua skew
fields. However, in U(V),
(1 + z)-'w(l + x) = (1 + z'Yxw'(\ + x')
so that
x = w-i(i + z)(i + z')-'w'(l
+ x') - I.
Hence Div(gp(x,z,w,x',z',w')) = Div(gp(z,w,x',z',w')) and U(V) does not
distinguish G from a free factor of G. However, it is clear by looking in a
Mal'cev-Neumann embedding of KG that if Hx and H2 are distinct subgroups of
G, then Div^g )(//,) j= T>iv,j(C)(H2).
Thus G Q U(V) is not a free embedding.
References
1. G. Baumslag,Positiveone relator groups,Trans. Amer. Math. Soc. 156 (1971),165-183.MR 43
#325.
2. P. M. Cohn,Freeidealrings,J. Algebra1 (1964),47-69;correctionibid.6 (1967),410.MR 28
#5095; 35 #214.
3.-,
Free rings and their relations, Academic Press, New York and London, 1971.
4. M. Dunwoody,Relationmodules,
Soc.4 (1972),151-155.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
346
JACQUESLEWIN
5.1. Hughes, Divisionrings offractionsfor group rings, Comm. Pure Appl. Math. 23 (1970), 181-188.
MR 41 #8533.
6. A. Karrass and D. Solitar, Onfinitely generated subgroupsof a free group, Proc. Amer. Math. Soc.
22 (1969),209-213. MR 39 #6961.
7.-,
The subgroups of a free product of two groups with an amalgamated subgroup, Trans. Amer.
Math. Soc. 150(1970),227-255.MR 41 #5499.
8. A. Klein, Involutorialdivisionrings with arbitrary center, Proc. Amer. Math. Soc. 34 (1972),38-42.
9. J. Lewin, A note on zero divisors in group rings, Proc. Amer. Math. Soc. 31 (1972), 357-359.
10. B. H. Neumann, On ordereddivisionrings.Trans. Amer. Math. Soc. 66 (1949),202-252. MR 11,
311.
Department of Mathematics, Syracuse University, Syracuse, New York 13210
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use